Help with Kirchhoff's Voltage Law math

I have always had a bit of trouble understanding KVL. In the attached image I need to measure all of the voltages in the two loops and confirm that
KVL works. I simply can not get the math on this one...can anyone help...the suggested solution is below?

Image: https://www.dropbox.com/s/jh3r82kxij9n5po/Screen Shot.png?dl=0

The solution is:
Use the voltage divider equation to get the voltage from node 1 to 3:
V = 12 (1 - 100/430) = 12 (0.768) = 9.22 V

Your problem is that you're trying to use the voltage divider equation with KVL. You can't do it like that because you then you don't account for some loops in the circuit (in this case you completely ignore the effect of 1K and 470 ohm resistors). You can't just go in and cherry pick the data you are interested in and pretend that the rest of the circuit has no effect.

What you have to do is draw current loops such paths in the circuit are accounted for (in this case loop 1 and 2, and then for each loop you follow the loop around while adding the voltage drops for all components in the loop (in this case they are all V = IR because only resistors) . You don't need a third loop equation for the center loop because all current paths for that loop have already been accounted for via loop 1 and 2.

Whenever two or more loops intersect a component, then the current flowing through that component is the sum of all currents in all the loops involved. The polarity used is whatever you defined as positive direction for each of the loops. It doesn't matter how you define each one, as long as you remain consistent.

For example, in this schematic:
https://www.elprocus.com/wp-content/uploads/2015/04/Example-Circuit-for-KVL-and-KCL.jpg

If the direction of the arrow in the loop is defined as positive then the equation for Loop 1 KVL is:

-Tracing in the direction of loop 1 arrow.
-Loop 2 arrow as drawn
0 = -10V + (R1*I1) + R3(I1 + I2)

-Tracing in the direction of loop 1 arrow.
-Loop 2 arrow drawn in the opposite direction
0 = -10V + (R1*I1) + R3(I1 - I2)

In the end it all works out to be the same thing once you take the direction of the drawn arrows into account.

To do KVL, you must always have equations that start off as 0 = sum of all voltages in each loop. That is what KVL is. Obviously, you have nothing of the sort with the voltage divider equation because the voltage divider equation is not KVL in it's most fundamental, general form. That means when you try to use the voltage divider equation you are making a lot of false assumptions about the circuit.

OK...I will give it a try...thanks so much.

dk: The 1K and 470 are directly across the source, There is nothing in the statement of the problem to indicate that the current out of the 12 V source is limited, so those resistors do not affect the voltages at nodes 1, 2, or 3 in any way.

GBB: The equation and solution are correct. What is it about the solution that you don't understand?

ak

AK - I don't understand (1 - 100/430)...where does the 1 come from?

AnalogKid said:

dk: The 1K and 470 are directly across the source, There is nothing in the statement of the problem to indicate that the current out of the 12 V source is limited, so those resistors do not affect the voltages at nodes 1, 2, or 3 in any way.

GBB: The equation and solution are correct. What is it about the solution that you don't understand?

ak

Click to expand...

Not if his goal is to verify KVL. The solution as written is pretty much the last line and skips all the actual work involving KVL. If he's trying to learn KVL, he has to to do the process I outlined and only in the very last line would he he end up with the solution he posted. If this was an exam wanted him to verify KVL works and he wrote just that solution out, he would get zero marks.

GBB said:

AK - I don't understand (1 - 100/430)...where does the 1 come from?

Click to expand...

Working backwards from the solution:
V = 12 (1 - 100/430)
V = 12[(430-100)/430]
V = 12[(330)/(100+330)]

->Resistive divider equation that you recognize but this isn't really recognizable as KVL anymore because by this point, all the currents have been cancelled out in the process of solving the solution. To understand KVL you need to start from the other end of the problem.

WHen I find the time I'll write down the full solution for you since it does get a bit tricky keeping track of everything.

Attached are two ways to do it. In all of these I follow around each current in the loop in the direction of the arrow (which I arbitrarily defined as positive) and whenever I ran into an current arrow pointing in the opposite direction of the current I am following, I use a negative sign). Keeping track of all the polarities is the most confusing part for beginners so draw them out.

In the first one, if you define 3 currents, then you have 3 unknowns and therefore need 3 equations to solve for all 3 currents. It is possible to completely solve this circuit by just defining I1 and I3 without an I2 since betwen those two current loops, all components and voltage drops are accounted for. That is the way your original schematic is drawn with the two lime green loops. However, you would need to use I2 if there were resistors in anywhere in the top or bottom horizontal wires in I1 or I2. You should try solving things the way it is drawn in your schematic with just a left and right current loop defined.

In the second one, instead of an independent current loop 3 being defined, a large super current loop is used instead. You can see from the simplicity of the equation for the big loop that depending on what info you are interested in, how you define your current loops can make things much faster or more direct to solve if you are only interested in a specific thing in your circuit.

Once you solve these equations for the currents you can calculate the voltage drop of component in the circuit (and you also know the current everywhere in the circuit. Remember that the current running through any part of the circuit is the sum of all the current loops running through that section.

The three steps to solving this: If you really want to apply KVL, apply it to the "simplified" version of the circuit so that you have two meshes. The DC analysis shows that the three versions of the circuit I show below are topologically equivalent.

Thanks for all of your help...you folks are great!

GBB said:

I have always had a bit of trouble understanding KVL. In the attached image I need to measure all of the voltages in the two loops and confirm that
KVL works. I simply can not get the math on this one...can anyone help...the suggested solution is below?

Image: https://www.dropbox.com/s/jh3r82kxij9n5po/Screen Shot.png?dl=0

The solution is:
Use the voltage divider equation to get the voltage from node 1 to 3:
V = 12 (1 - 100/430) = 12 (0.768) = 9.22 V

Click to expand...

Kirchhoff's voltage law (KVL) states that the voltage drops across each component in a circuit loop are equal to the source voltage. You do not need the voltage divider rule to validate KVL. Only the rightmost loop with the two resistors are relevant to this problem. Using the definition of resistance, it can be determined that 12/(330+100) =0 .027907 amps of current are present in that branch. Since the resistors are in series, the same current is present in both resistors. So if we figure out the voltage drop across each resistor and add the voltages, we should get a total of 12 volts, right? The voltage across R100 is 100*0.027907 = 2.7907 volts. The voltage across R330 is 330*0.027907 = 9.2093 volts. Adding the voltages of the two resistors we get 12 volts. That shows the sum the voltage drops across the two resistors is equal to the source voltage and validates KVL. Any questions?

Ratch

Notice: This validates KVL, but it does not prove it. We used specific values of voltage and resistance. A real proof would use a general voltage and components. Otherwise, a pedantic person could say that it might not work for a different set of values.

Watch - thanks so much for your explanation.